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I have written a proof for the following and would like you to correct me if I made any mistakes, thanks in advance:

Claim: $X = \sqcup_i X_i$ then $H_q (X) \cong \oplus_i H_q (X_i)$

proof: Proof of case $X = A \sqcup B$, the general case follows by induction.

By the Mayer-Vietoris theorem the following sequence is exact:

$$ \dots \xrightarrow{k_\ast} H_{n+1}(X) \xrightarrow{\partial_\ast} H_n(A \cap B) \xrightarrow{(i_\ast, j_\ast)} H_n(A) \oplus H_n(B) \xrightarrow{k_\ast} H_n(X)\xrightarrow{\partial_\ast} \dots$$

Then $A \cap B = \emptyset \implies H_n(A \cap B) = 0 \implies \partial_\ast = 0$

Then $k_\ast$ is injective because $ker k_\ast = im \partial_\ast = 0$ and $k_\ast $ is surjective because $im k_\ast = ker \partial_\ast = H_n(X)$

Arctic Char
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Rudy the Reindeer
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    Well, yes, if you take Mayer-Vietoris for granted. However, you should probably try and prove this directly from the definition of homology... Suggestion: go through Hatcher's proof of Mayer-Vietoris and extract a direct argument. This would probably be more illuminating than what you did. – t.b. Aug 08 '11 at 11:39
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    If your collection is not countable, you will need transfinite induction. – gary Aug 08 '11 at 11:42
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    Maybe you can argue that if a cycle bounds in one of the $A_i$'s, then it will bound in the disjoint union, and, conversely, a trivial cycle will also be trivial in the union. – gary Aug 08 '11 at 12:09
  • @Theo, ok, I'll do that. – Rudy the Reindeer Aug 08 '11 at 12:11
  • @gary, hey thanks for the hint! – Rudy the Reindeer Aug 08 '11 at 12:12
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    You could also make an analogous observation about the singular (or simplicial, or $\Delta$) chain complex for the disjoint union and work from there. – wckronholm Aug 08 '11 at 15:54
  • @wckronholm: thanks for the hint! – Rudy the Reindeer Aug 08 '11 at 19:17
  • @MattN. you can tell me how to solve now you have the answer, please, i also have a question, $A$ and $B$ are closed so how you can use the Mayer-Vietoris theorem ? Thank you – Vrouvrou Aug 31 '14 at 08:28
  • @Vrouvrou Sorry this is way too long ago and I think I never really understood Mayer-Vietoris, sadly. – Rudy the Reindeer Sep 01 '14 at 07:01

3 Answers3

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Wouldn't be easier to argue like this?

Since $\Delta^n$ is connected, then the image of every continuous map $\sigma : \Delta^n \longrightarrow \bigsqcup_{\alpha \in J} X_\alpha$ must be contained in some $X_\alpha$: $\sigma (\Delta^n) \subset X_\alpha$.

I didn't check the details, but I think that this would say that you can find an inverse to the universal map

$$ \bigoplus_{\alpha \in J} H_p(X_\alpha) \longrightarrow H_p(\bigsqcup_{\alpha \in J} X_\alpha) $$

induced by the inclusions $X_\alpha \longrightarrow \bigsqcup_{\alpha \in J} X_\alpha$. And this would be true for any set of indices $J$.

Agustí Roig
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The important thing is to understand where the finiteness comes from. By Hatcher Proposition A.1, a compact subspace of a CW complex is contained in a finite subcomplex. Since the image of a union of $\Delta^k$ in $X$ is a compact subspace, it is contained in a finite subcomplex $\bigoplus_{i=1}^mC_k(X_{\alpha_i})$, so the inclusion from $\bigoplus_\alpha H_k(X_\alpha)$ is surjective, hence the isomorphism.

Dasheng Wang
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Let $G$ be an abelian group and $G=\oplus_\alpha G_\alpha$. For each $\alpha$, let $H_\alpha$ be a subgroup of $G_\alpha$. Let $H=\oplus_\alpha H_\alpha$. For each $\alpha$, let $q_\alpha:G_\alpha\to G_\alpha/H_\alpha$ be the quotient map. We get a map $\oplus_\alpha q_\alpha:\oplus_\alpha G_\alpha\to \oplus_\alpha G_\alpha/H_\alpha$. The map $\oplus_\alpha q_\alpha$ is onto and it's kernel is $H$. So we get that $G/H\simeq \oplus_\alpha G_\alpha/H_\alpha$.

We will now state a general result. Given an arbitrary collection of chain complexes of abelian groups $\{(A^\alpha_\bullet,\partial_\bullet^\alpha):\alpha\in\Lambda\}$ we can form the chain complex $(A_\bullet,\partial_\bullet)$ where for each $n\in\mathbb Z$, we define $A_n=\oplus_\alpha A^\alpha_n$ and $\partial_n=\oplus_\alpha \partial^\alpha_n$. We can check that the homology groups of this complex $H_n(A_\bullet)\simeq \oplus_\alpha H_n(A^\alpha_\bullet)$.

Suppose $\{X_\alpha:\alpha\in\Lambda\}$ are the path-components of a space $X$. For each $n\in\mathbb N$, let $C_n(X)$ be the group of singular $n$-chains in $X$. Let $\Delta^n$ denote the standard $n$-simplex and let $\sigma:\Delta^n\to X$ be a singular $n$-simplex. As $\Delta^n$ is path-connected, so is the image of $\sigma$ and hence it is contained in some path-component of $X$. As any singular $n$-chain is a finite formal $\mathbb Z$-linear combination of singular $n$-simplices, it follows that $C_n(X)=\oplus_\alpha C_n(X_\alpha)$.

From the definition of the boundary map $\partial_n:C_n(X)\to C_{n-1}(X)$ and the boundary maps $\partial_n^\alpha:C_n(X_\alpha)\to C_{n-1}(X_\alpha)$, it follows that $\partial_n^\alpha$ is obtained by the restriction of the map $\partial_n$ to $C_n(X_\alpha)$ for each $\alpha\in\Lambda$. Thus, we get that $\partial_n=\oplus_\alpha \partial_n^\alpha$. Hence, from the above mentioned general result, we see that $H_n(X)=\oplus_\alpha H_n(X_\alpha)$.

Shameek Paul
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